Problem: Simplify and expand the following expression: $ \dfrac{3}{z - 7}- \dfrac{5}{4z + 32}+ \dfrac{z}{z^2 + z - 56} $
First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $4$ out of denominator in the second term: $ \dfrac{5}{4z + 32} = \dfrac{5}{4(z + 8)}$ We can factor the quadratic in the third term: $ \dfrac{z}{z^2 + z - 56} = \dfrac{z}{(z - 7)(z + 8)}$ Now we have: $ \dfrac{3}{z - 7}- \dfrac{5}{4(z + 8)}+ \dfrac{z}{(z - 7)(z + 8)} $ The least common multiple of the denominators is: $ (z - 7)(z + 8)$ In order to get the first term over $(z - 7)(z + 8)$ , multiply by $\dfrac{4(z + 8)}{4(z + 8)}$ $ \dfrac{3}{z - 7} \times \dfrac{4(z + 8)}{4(z + 8)} = \dfrac{12(z + 8)}{(z - 7)(z + 8)} $ In order to get the second term over $(z - 7)(z + 8)$ , multiply by $\dfrac{z - 7}{z - 7}$ $ \dfrac{5}{4(z + 8)} \times \dfrac{z - 7}{z - 7} = \dfrac{5(z - 7)}{(z - 7)(z + 8)} $ In order to get the third term over $(z - 7)(z + 8)$ , multiply by $\dfrac{4}{4}$ $ \dfrac{z}{(z - 7)(z + 8)} \times \dfrac{4}{4} = \dfrac{4z}{(z - 7)(z + 8)} $ Now we have: $ \dfrac{12(z + 8)}{(z - 7)(z + 8)} - \dfrac{5(z - 7)}{(z - 7)(z + 8)} + \dfrac{4z}{(z - 7)(z + 8)} $ $ = \dfrac{ 12(z + 8) - 5(z - 7) + 4z} {(z - 7)(z + 8)} $ Expand: $ = \dfrac{12z + 96 - 5z + 35 + 4z}{4z^2 + 4z - 224} $ $ = \dfrac{11z + 131}{4z^2 + 4z - 224}$